Complex equation The problem says to solve the given complex equation: 
$$
z^4-\left[
\frac{\sqrt3}{2}i^{21}+
\frac{\sqrt3}{2}i^{9}+
\frac{8}{(1+i)^6}\right]^9=0
$$
The solution is this:
$$
\cos\left(
\frac{\pi}{8}+\frac{k\pi}{2}
\right)+i\sin\left(
\frac{\pi}{8}+\frac{k\pi}{2}
\right)
,k=0,1,2,3
$$
My problem is that I can't get this solution, I've tried multiple times to solve it but I always end up getting this solution:
$$z^4=(i+\sqrt3*i)^9$$
Could you help me with this? Thanks in advance.
 A: I believe that the work you did is correct, but the solution is somewhat wrong. Here's how I started solving the equation:
$$
z^4-\left[
\frac{\sqrt3}{2}i^{21}+
\frac{\sqrt3}{2}i^{9}+
\frac{8}{(1+i)^6}
\right]^9=0
\\
z^4=\left[
\frac{\sqrt3}{2}i^{21}+
\frac{\sqrt3}{2}i^{9}+
\frac{8}{(1+i)^6}
\right]^9
\\
z^4=\left[
\frac{\sqrt3}{2}i+
\frac{\sqrt3}{2}i+
\frac{8}{-8i}
\right]^9
\\
z^4=(i+i\sqrt3)^9
$$
According to this, your work is correct. In order to finish solving this equation, we need to find all four fourth roots of the number $(i+i\sqrt3)^9$.
The simplest way to calculate all of the roots of a complex number is to convert it to polar form, and then use De Moivre's formula to calculate the roots in polar form.
We convert the number $(i+i\sqrt3)^9$ to polar form:
$$
(i+i\sqrt3)^9=
i(4240+2448\sqrt3)
=
(4240+2448\sqrt3)
\left[
\cos\frac{\pi}{2}
+i\sin\frac{\pi}{2}
\right]
$$
By using De Moivre's formula, we can arrive to the following solution:
$$
z\in \left\{
\sqrt[4]{4240+2448\sqrt3}
\left[
\cos\left(
\frac{\pi}{8}+\frac{k\pi}{2}
\right)+i\sin\left(
\frac{\pi}{8}+\frac{k\pi}{2}
\right)
\right]
,k=0,1,2,3
\right\}
$$
As you can see, the solution you provided is similar to the correct one, but lacks the absolute value part in the solution, which is essential when writing a complex number in polar form.
You should probably notify whoever you got this solution from about the error.
A: $$z^4-\left(\frac{\sqrt3}{2}i^{21}+\frac{\sqrt3}{2}i^{9}+\frac{8}{(1+i)^6}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{((1+i)^3)^2}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{((1+i)(1+i)^2)^2}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{((1+i)2i)^2}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{2^2\cdot -2i}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{-8i}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+i\right)^9=0\Longleftrightarrow$$
$$z^4-\left(i(1+\sqrt{3})\right)^9=0\Longleftrightarrow$$
$$z^4-\left(i(1+\sqrt{3})\right)^9=0\Longleftrightarrow$$
$$z^4=\left(i(1+\sqrt{3})\right)^9\Longleftrightarrow$$
$$z^4=\left|\left(i(1+\sqrt{3})\right)^9\right|\cdot e^{\arg\left(\left(i(1+\sqrt{3})\right)^9\right)i}\Longleftrightarrow$$
$$z^4=(1+\sqrt{3})^9\cdot e^{\frac{\pi}{2}i}\Longleftrightarrow$$
$$z=\left((1+\sqrt{3})^9\cdot e^{\left(\frac{\pi}{2}+2\pi k\right)i}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$z=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi k\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-3$

So the solutions are:
$$z_0=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 0\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
$$z_1=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 1\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
$$z_2=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 2\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
$$z_3=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 3\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
